Hi
(8x^2 + 1)/(2x^2-5x-3) |cannot divide by ZERO
2x^2 -5x - 3 = 0
factoring
(2x +1)(x-3) = 0
(2x +1)=0 x = -1/2
(x-3) = 0 x = 3
{x: x ≠ 3 and x ≠ -1/2} the set of all real numbers except 3 and -1/2
Use the "^" character (Shift+6) to indicate exponents.
Put multiple term numerators and denominators in parentheses
Do not try to post fractions vertically like you did. Instead separate the numerator and denominator by a "/" character.
Here's your expression written accordin to these guidelines:
(8x^2+1)/(2x^2-5x-3)
Problems that are clearly posted are more like to get a response.
To find a domain you start by assuming that the domain is all real numbers. Then you start ruling out values, if any, that would create an expression you should never have. As you go through Math you learn of things that should not happen. Probably the first one you learn of si dividing by zero. We can never have an expression that has a division by zero! (There are other things to avoid (square roots of negative numbers, etc.) but none of them apply to your expression.)
Your expression has a denominator. So we must limit the domain so that the denominator is never zero. So we have to find out what numbers would make the denominator zero and then rule out those numbers. To find the numbers that make the denominator zero, you set the denominator to zero:
and solve the equation.
This is a quadratic equation so we factor it (or use the Quadratic Formula). This factors:
(2x+1)(x-3) = 0
From the Zero Product Property we know that one of these factors must be zero:
2x+3 = 0 or x-3 = 0
Solving these we get:
x = -3/2 or x = 3
So the domain is all real numbers except these two! Expressing this in set notation is difficult here because the proper symbols are hard to type. The best I can do is:
{x : and }
Instead of the word "and" you should use the upside-down U that is set notation symbol for intersection/and. Also, some places use